Question

If in a $\Delta ABC,\angle C={90}^0$, then the maximum value of cos$A$ cos$B$ is:

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $\frac{1}{2}$

$cosAcosB=\frac{1}{2}\left[cos\right(A+B)+cos(A-B\left)\right]$
Given $\angle C={90}^o,\angle A+\angle B={90}^0$
$cosAcosB=\frac{1}{2}[0+cos(A-B\left)\right]$
$\therefore$for max. $cosAcosB,cos(A-B)=1$
$\therefore cosAcosB=\frac{1}{2}$